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Basic Math Question (matrices)

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that the matrix B= 4 -2 -2
    1 1 -1
    2 -2 0
    satisfies the quadratic equation
    B^2-3B+2I=0

    Hence show that B^3=7B-6I


    3. The attempt at a solution
    Now i have already worked out the quadratic equation and the matirx B satisfies the equaition B^2-3B+2I=0.

    My problem is i cant seem to prove the next part i am stuck here is how for i got


    B^3=7B-6I
    multiplying by B^2 gives

    B^4 - 3B^3 + 2B^2I = O

    multiply b B^-1 (B inverse) gives

    IB^3 - 3IB^2 = 2IB =0

    this is where i am stuck.. thanks for any help given...question dued tomorrow.
     
  2. jcsd
  3. Nov 19, 2008 #2

    Mark44

    Staff: Mentor

    You said you showed that B satisfies B^2 - 3B + 2I = 0. Now you want to show that the same matrix satisfies B^3 - 7B + 6I = 0.

    These matrix equations work much like the equations you worked with earlier in your math classes with regard to factorization. For example, your first equation factors into
    (B - 2I)(B - I) = 0. Instead of doing what you did before (multiplying B by itself, subtracting 3*B, and then adding 2*I), you could have calculation B - 2I and B - I and multiplied these two matrices together.

    As it turns out, your second equation can also be factored, and one of the factors is B^2 - 3B + 2I.
     
  4. Nov 19, 2008 #3

    HallsofIvy

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    You are given B so just multiply to get B3. Then calculuate 7B- 6I and see if they are the same!
     
  5. Nov 19, 2008 #4
    yes i agree but we cant do it that we we have to use the quadratic equation to prove that B^3 = 7B-6I
     
  6. Nov 19, 2008 #5

    HallsofIvy

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    Then do the obvious thing: divide x3- 7x+ 6 by x2- 3x+ 2.
     
  7. Nov 19, 2008 #6
    B^2=3b-2i
    b^3=3b^2-2b=3(3b-2i)-2b=7b-6i
     
  8. Nov 19, 2008 #7
    and how will that help me i get x+3 :|
     
  9. Nov 19, 2008 #8
    Thanks alot cloudyhill that helps greatly!
     
  10. Nov 19, 2008 #9

    gabbagabbahey

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    That tells you that [itex]B^3-7B+6I=(B+3I)(B^2-3B+2I)[/itex] but you already showed that [itex]B^2-3B+2I=0[/itex], sooo [itex]B^3-7B+6I=\ldots[/itex]?:wink:
     
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